ACM思维题训练集合 You are given two integers n and d. You need to construct a rooted binary tree consisting of n vertices with a root at the vertex 1 and the sum of depths of all vertices equals to d. A tree is a connected graph without cycles. A rooted tre…

Solution 预处理出 \(i\) 个点组成的二叉树的最大答案和最小答案 递归做,由于只需要构造一种方案,我们让左子树大小能小就小,因此每次从小到大枚举左子树的点数并检验,如果检验通过就选定之 现在还需要确定左右子树各分配多少答案上去,一种构造性的想法是,要么让左子树选最小,要么让右子树选最大.对于任意一种其它方案,我们可以通过把左子树上的答案不断移到右子树上,直到某一边达到界限,故等价. 打错变量名查半天-- #include <bits/stdc++.h> using namespac…

Binary Tree 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5573 Description The Old Frog King lives on the root of an infinite tree. According to the law, each node should connect to exactly two nodes on the next level, forming a full binary tree. S…

function createNode(value) { return { value, left: null, right: null }; } function BinaryTree(val) { return { root: null, nodes: [], add(val) { const node = createNode(val); if (!this.root) { this.root = node; } else { this.downShift(node); } this.no…

题意:给定节点数n和所有节点的深度总和d,问能否构造出这样的二叉树.能,则输出“YES”,并且输出n-1个节点的父节点(节点1为根节点). 题解:n个节点构成的二叉树中,完全(满)二叉树的深度总和最小,单链树(左/右偏数)的深度总和最大.若d在这个范围内,则一定能构造出来:否则一定构造不出来. 1.初始构造一颗单链树,依次把底部的节点放入上面的层,直到满足深度总和为d 2.若当前深度总和sum > d,则先拿掉底端节点. 拿掉后,若sum依然比d大,就直接把底端节点放入有空位的最上层: 拿掉后s…

膜这场比赛的 \(rk1\) \(\color{black}A\color{red}{lex\_Wei}\) 这题应该是这场比赛最难的题了 容易发现,二叉树的下一层不会超过这一层的 \(2\) 倍,所以我们先构造出来一颗尽量满的二叉树,然后慢慢向下调整,调整的方法是从最上面一个一个弄下来. 然后你慢慢调整的复杂度最多是 \(d\) ,复杂度 \(O(d)\) #include <bits/stdc++.h> using namespace std ; const int maxn = 5e3…

http://www.geeksforgeeks.org/full-and-complete-binary-tree-from-given-preorder-and-postorder-traversals/ #include <iostream> #include <vector> #include <algorithm> #include <queue> #include <stack> #include <string> #in…

Given preorder and inorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. For example, given preorder = [3,9,20,15,7] inorder = [9,3,15,20,7] Return the following binary tree: 3 / \ 9 20…

Given inorder and postorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. For example, given inorder = [9,3,15,20,7] postorder = [9,15,7,20,3] Return the following binary tree: 3 / \ 9 2…

Given preorder and inorder traversal of a tree, construct the binary tree. Note:  You may assume that duplicates do not exist in the tree. 利用前序和中序遍历构造二叉树的思想与利用中序和后续遍历的思想一样,不同的是,全树的根节点为preorder数组的第一个元素,具体分析过程参照利用中序和后续构造二叉树.这两道题是从二叉树的前序遍历.中序遍历.后续遍历这三种遍…

Given inorder and postorder traversal of a tree, construct the binary tree. Note:  You may assume that duplicates do not exist in the tree. 利用中序和后序遍历构造二叉树,要注意到后序遍历的最后一个元素是二叉树的根节点,而中序遍历中,根节点前面为左子树节点后面为右子树的节点.例如二叉树:{1,2,3,4,5,6,#}的后序遍历为4->5->2->6-&…

Given preorder and inorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree.从前序以及中序的结果中构造二叉树,这里保证了不会有两个相同的数字,用递归构造就比较方便了: class Solution { public: TreeNode* buildTree(vector<int>& preor…

题目: Given inorder and postorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. 说明: 1)实现与根据先序和中序遍历构造二叉树相似,题目参考请进 算法思想 中序序列:C.B.E.D.F.A.H.G.J.I   后序序列:C.E.F.D.B.H.J.I.G.A   递归思路: 根据后序遍历的特点,…

题目: Given preorder and inorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. 说明: 1)二叉树可空 2)思路:a.根据前序遍历的特点, 知前序序列(PreSequence)的首个元素(PreSequence[0])为二叉树的根(root),  然后在中序序列(InSequence)中查找此根(…

我们都知道,已知中序和后序的序列是可以唯一确定一个二叉树的. 初始化时候二叉树为:================== 中序遍历序列,           ======O=========== 后序遍历序列,           =================O 红色部分是左子树,黑色部分是右子树,O是根节点 如上图所示,O是根节点,由后序遍历可知, 根据这个O可以把找到其在中序遍历当中的位置,进而,知道当前这个根节点O的左子树的前序遍历和中序遍历序列的范围. 以及右子树的前序遍历和中序遍历…

我们都知道,已知前序和中序的序列是可以唯一确定一个二叉树的. 初始化时候二叉树为:================== 前序遍历序列,           O================= 中序遍历序列,           ======O=========== 红色部分是左子树,黑色部分是右子树,O是根节点 如上图所示,O是根节点,由前序遍历可知, 根据这个O可以把找到其在中序遍历当中的位置,进而,知道当前这个根节点O的左子树的前序遍历和中序遍历序列的范围. 以及右子树的前序遍历和中序遍历…

Given inorder and postorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree.同样是给出了不会有重复数字的条件,用递归较容易实现,代码如下: /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNod…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 递归 日期 题目地址:https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/description/ 题目描述 Given preorder and inorder traversal of a tree, construct t…

Construct Binary Tree from Preorder and Inorder Traversal Given preorder and inorder traversal of a tree, construct the binary tree. 分析: 根据前序遍历和中序遍历构造一棵树,递归求解即可 /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left;…

原题地址: https://oj.leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/ 题目内容: Given preorder and inorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. 方法: 非常单纯的…

一天一道LeetCode 本系列文章已全部上传至我的github,地址:ZeeCoder's Github 欢迎大家关注我的新浪微博,我的新浪微博 欢迎转载,转载请注明出处 (一)题目 来源:https://leetcode.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/ Given inorder and postorder traversal of a tree, construct the b…

一天一道LeetCode 本系列文章已全部上传至我的github,地址:ZeeCoder's Github 欢迎大家关注我的新浪微博,我的新浪微博 欢迎转载,转载请注明出处 (一)题目 来源:https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/ Given preorder and inorder traversal of a tree, construct the bin…

Construct Binary Tree from Preorder and Inorder Traversal Given preorder and inorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. SOLUTION 1: 1. Find the root node from the preorder.(it…

Given preorder and inorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. 难度:95,参考了网上的思路.这道题是树中比较有难度的题目,需要根据先序遍历和中序遍历来构造出树来.这道题看似毫无头绪,其实梳理一下还是有章可循的.下面我们就用一个例子来解释如何构造出树.假设树的先序遍历是12453687,…

Given inorder and postorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. 解题思路: 给出一个二叉树的中序和后序遍历结果,还原这个二叉树. 对于一个二叉树: 1 / \ 2 3 / \ / \ 4 5 6 7 后序遍历结果为:4 5 2 6 7 3 1 中序遍历结果为:4 2 5 1 6 3 7…

Given preorder and inorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. 解题思路: 给出一个二叉树的先序和中序遍历结果,还原这个二叉树. 对于一个二叉树: 1 / \ 2 3 / \ / \ 4 5 6 7 先序遍历结果为:1 2 4 5 3 6 7 中序遍历结果为:4 2 5 1 6 3 7 由…

Given inorder and postorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. 题解:如下图所示的一棵树: 5 / \ 2 4 / \ \ 1 3 6 中序遍历序列:1  2  3  5  4  6 后序遍历序列:1  3  2  6  4  5 后序遍历序列的最后一个元素就是当前根节点元素.首先想到的…

题目 Given inorder and postorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. Show Tags Show Similar Problems 分析 跟上一道题同样的道理. AC代码 class Solution { public: template <typename Iter> TreeN…

题目 Given preorder and inorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. 分析 给定一颗二叉树的前序和中序遍历序列,求该二叉树. 我们手动做过很多这样的题目,掌握了其规则~ 前序遍历第一个元素为树的root节点,然后在中序序列中查找该值,元素左侧为左子树,右侧为右子树: 求出左子树个数cou…

Given inorder and postorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. 这道题要求从中序和后序遍历的结果来重建原二叉树,我们知道中序的遍历顺序是左-根-右,后序的顺序是左-右-根,对于这种树的重建一般都是采用递归来做,可参见我之前的一篇博客Convert Sorted Array to Bin…